Large Deviations, Dynamics and Phase Transitions in Large Stochastic and Disordered Neural Networks

Abstract

Neuronal networks are characterized by highly heterogeneous connectivity, and this disorder was recently related experimentally to qualitative properties of the network. The motivation of this paper is to mathematically analyze the role of these disordered connectivities on the large-scale properties of neuronal networks. To this end, we analyze here large-scale limit behaviors of neural networks including, for biological relevance, multiple populations, random connectivities and interaction delays. Due to the randomness of the connectivity, usual mean-field methods (e.g. coupling) cannot be applied, but, similarly to studies developed for spin glasses, we will show that the sequences of empirical measures satisfy a large deviation principle, and converge towards a self-consistent non-Markovian process. From a mathematical viewpoint, the proof differs from previous works in that we are working in infinite-dimensional spaces (interaction delays) and consider multiple cell types. The limit obtained formally characterizes the macroscopic behavior of the network. We propose a dynamical systems approach in order to address the qualitative nature of the solutions of these very complex equations, and apply this methodology to three instances in order to show how non-centered coefficients, interaction delays and multiple populations networks are affected by disorder levels. We identify a number of phase transitions in such systems upon changes in delays, connectivity patterns and dispersion, and particularly focus on the emergence of non-equilibrium states involving synchronized oscillations.

Appendices

Appendix A: H is a Good Rate Function

This appendix is devoted to the proof of the Proposition 3 stating that the map H given by (20) is a good rate function, namely that it is lower semi-continuous with compact level sets. In order to perform this demonstration, it is convenient to analyze for a moment an intermediate system where the interaction is discretized in time, which will expedite the analysis of our continuous time problem.

Given an integer k, we define Δ^k={0=t ₀<t ₁<⋯<t _k <t _k+1=T} a partition of [0,T] and consider the following dynamics for the neurons of population α:

$$\begin{aligned} \begin{cases} dx^i_t= (-\frac{1}{\theta_{\alpha}}x^i_t + \sum_{\gamma=1}^M \sum_{j: p(j)=\gamma} J_{ij} S_{\alpha p(j)}(x^j_{t^{(k)}-\tau_{\alpha p(j)}}) )\,dt+ \lambda_{\alpha}dW^{i}_t\\ t^{(k)}= \sup\{ t_l \in\Delta^k | t_l \leq t \}\\ \mbox{Law of } (x_t^{\alpha})_{t\in[-\tau,0]} = \mu_{\alpha }^{\otimes N_{\alpha}}. \end{cases} \end{aligned}$$

(33)

As in Proposition 1, this system clearly admits a unique weak solution for any $J \in\mathbb{R}^{N \times N}$. We will denote Q ^n,α,(k)(J) its restriction to the σ-algebra $\sigma(x^{i}_{s}, 1\leq i \leq N ; p(i)=\alpha, -\tau\leq s\leq T)$, and $Q^{n,\alpha,(k)}=\mathcal{E}(Q^{n,\alpha,(k)}(J))$. They are both probability measures on $\mathcal{C}([-\tau,T],\mathbb{R})^{n}$. By Girsanov Theorem, $Q^{n,\alpha,(k)}(J) \ll P_{\alpha}^{\otimes n}$ with:

$$\begin{aligned} \frac {\mbox {d} Q^{n,\alpha,(k)}(J)}{\mbox {d} {P_{\alpha}}^{\otimes n}} =& \exp\Biggl\{ \sum_{i:p(i)=\alpha} \int_{0}^{T} \Biggl(\frac{1}{\lambda_{\alpha}} \sum_{\gamma=1}^{P} \sum_{j:p(j)=\gamma} J_{ij} S_{\alpha\gamma} \bigl(x_{t^{(k)}-\tau _{\alpha\gamma}}^j\bigr) \Biggr) dW_{t}^{i} \\ &{}- \int_{0}^{T} \Biggl(\frac{1}{\lambda_{\alpha}} \sum _{\gamma =1}^{P} \sum _{j:p(j)=\gamma} J_{ij} S_{\alpha\gamma}\bigl(x_{t^{(k)}-\tau_{\alpha \gamma}}^j \bigr) \Biggr)^2 dt \Biggr\}. \end{aligned}$$

We define for $\mu\in\mathcal{M}_{1}^{+}(\mathcal{C})$ the discrete-time version of Γ(μ), denoted Γ ^k(μ) on Δ^k, as:

$$\begin{aligned} \varGamma^{k}(\mu) =& \int_{\mathcal{C}} \log\Biggl( \int \exp\Biggl\{ \sum_{l=0}^k \bigl( \mathbf{G}_{t_l}(\omega)+\mathbf{m}_{\mu}(t_l) \bigr)' \cdot(\mathbf{W} _{t_{l+1}} - \mathbf{W} _{t_l} ) (x) \\ &{}- \frac{1}{2} \sum_{l=0}^k \bigl\| \mathbf{G}_{t_l}(\omega)+\mathbf{m} _{\mu}(t_l) \bigr\|^2(t_{l+1}-t_l) \Biggr\} d \gamma_{K_{\mu}}(\omega) \Biggr) d\mu(x). \end{aligned}$$

Since the covariance K _μ is diagonal and therefore the components of G under γ _μ are independent, this map is the sum of the functions Γ ^α,k given by:

$$\begin{aligned} \varGamma^{\alpha,k}(\mu) =& \int_{\mathcal{C}} \log\Biggl( \int \exp\Biggl\{ \sum_{l=0}^k \bigl(G_{t_l}^{\alpha}(\omega)+m_{\mu}^{\alpha}(t_l) \bigr) \bigl(W^{\alpha}_{t_{k+1}} - W^{\alpha}_{t_l} \bigr) (x) \\ &{}- \frac{1}{2} \sum_{l=0}^k \bigl(G_{t_l}^{\alpha}(\omega)+m_{\mu}^{\alpha}(t_l) \bigr)^2(\omega) (t_{l+1}-t_l) \Biggr\} d \gamma_{K_{\mu}}(\omega) \Biggr) d\mu(x). \end{aligned}$$

The discretized version of our function H of interest is simply:

$$\begin{aligned} H^k (\mu) = \left\{ \begin{array}{l@{\quad}l} I(\mu|P) - \varGamma^k(\mu) & \mbox{if } I(\mu|P) < \infty,\\ \infty& \mbox{otherwise}. \end{array} \right. \end{aligned}$$

We will show that the map H ^k is a good rate function. This proof proceeds by introducing additional maps on $\mathcal{M}_{1}^{+}(\mathcal{C})$:

$$\begin{aligned} \varGamma_1^k(\mu) =& \log\biggl( \int\exp\biggl( - \frac{1}{2} \int_0^T \bigl\| \mathbf{G}_{t^{(k)}}(\omega) \bigr\|^2dt \biggr) d\gamma_{K_{\mu }}(\omega) \biggr) - \frac{1}{2} \int _0^T \bigl\|\mathbf{m}_{\mu} \bigl(t^{(k)}\bigr) \bigr\|^2 dt\\ \varGamma_2^k(\mu) =& \frac{1}{2} \int\int\biggl( \int_0^T \mathbf{G}_{t^{(k)} }' \cdot d\mathbf{W}_t(x) - \mathbf{m}_{\mu} \bigl(t^{(k)}\bigr)dt\biggr) ^2 d\gamma_{\widetilde {K}_{\mu}^{T,k}} d \mu(x) \\ &{}+\int\int\mathbf{m}_{\mu}\bigl(t^{(k)}\bigr)' \cdot d\mathbf{W}_t (x) d\mu(x) \end{aligned}$$

where

$$\begin{aligned} &\widetilde{K}_{\mu}^{t,k}(s,u)\\ &\quad{}= \biggl(\int\frac{ \exp\{ - \frac {1}{2} \int_0^t (G^{\alpha}_{u^{(k)}}(\omega) )^2+\mathbf {1}_{\alpha \neq\gamma} (G^{\gamma}_{u^{(k)}}(\omega) )^2du \} G^{\gamma}_{u^{(k)} }(\omega)G^{\alpha}_{s^{(k)}}(\omega) }{\int\exp\{ - \frac {1}{2} \int_0^t (G^{\alpha}_{u^{(k)}}(\omega) )^2+\mathbf{1}_{\alpha\neq \gamma } (G^{\gamma}_{u^{(k)}}(\omega) )^2du \} d\gamma_{\mu}} d\gamma_{\mu} \biggr)_{\alpha, \gamma\in\{1 \cdots M\}}. \end{aligned}$$

One can easily see that this function takes values in the M×M diagonal positive matrices. Moreover, we can define $\gamma _{\widetilde {K}_{\mu}^{T,k}}$, probability measure on Ω, such that

$$\begin{aligned} d\gamma_{\widetilde{K}_{\mu}^{T,k}}= \frac{\prod_{\alpha=1}^M \exp \{ - \frac{1}{2} \int_0^T (G^{\alpha}_{t^{(k)}}(\omega) )^2 dt \} }{\int\prod_{\alpha=1}^M \exp\{ - \frac{1}{2} \int_0^T (G^{\alpha}_{t^{(k)}}(\omega) )^2 dt \} d\gamma_{\mu}} d\gamma_{\mu}, \end{aligned}$$

under which G is a M-dimensional centered Gaussian process with covariance $\widetilde{K}_{\mu}^{T,k}$ (this Gaussian calculus property is proved for instance in [3, Appendix A]).

Proposition 4

$$\begin{aligned} \varGamma^{k}(\mu)= \varGamma_1^{k}(\mu) + \varGamma_2^{k}(\mu) \end{aligned}$$

Proof

Let

$$\begin{aligned} \varGamma_1^{\alpha,k}(\mu) =& \log\biggl( \int\exp\biggl( - \frac {1}{2} \int_0^T {G^{\alpha}_{t^{(k)}}}^2( \omega)dt \biggr) d\gamma_{K_{\mu }}(\omega) \biggr) - \frac{1}{2} \int _0^T \bigl(m_{\mu}^{\alpha} \bigl(t^{(k)}\bigr)\bigr)^2 dt,\\ \varGamma_2^{\alpha,k}(\mu) =& \frac{1}{2} \int\int\biggl( \int_0^T G_{t^{(k)} }^{\alpha} \bigl(dW_t^{\alpha}(x) - m^{\alpha}_{\mu} \bigl(t^{(k)}\bigr)dt\bigr) \biggr)^2 d\gamma _{\widetilde{K}_{\mu}^{T,k}} d\mu(x)\\ &{} + \int\int m_{\mu}^{\alpha }\bigl(t^{(k)} \bigr)dW^{\alpha}_t (x) d\mu(x). \end{aligned}$$

For $\varGamma_{i}^{k}= \sum_{\alpha=1}^{M} \varGamma_{i}^{\alpha,k}, i\in \{1,2\}$, it is sufficient to prove that

$$\begin{aligned} \varGamma^{\alpha,k}(\mu)= \varGamma_1^{\alpha,k}(\mu) + \varGamma_2^{\alpha ,k}(\mu). \end{aligned}$$

Simple manipulations yield

$$\begin{aligned} \varGamma^{\alpha, k}(\mu) = & \int\log\biggl( \int\exp\biggl\{ \int _0^T \bigl(G^{\alpha}_{t^{(k)}}( \omega)+m_{\mu}^{\alpha}\bigl(t^{(k)}\bigr) \bigr)dW_t^{\alpha }(x) \\ &{} - \frac{1}{2} \int_0^T \bigl(G_{t^{(k)}}^{\alpha}(\omega)+m_{\mu }^{\alpha } \bigl(t^{(k)}\bigr) \bigr)^2 dt \biggr\} d \gamma_{K_{\mu}}(\omega) \biggr) d\mu(x) \\ = & \int\log\biggl\{ \biggl( \exp\biggl\{-\frac{1}{2} \int _0^T \bigl(m^{\alpha }_{\mu} \bigl(t^{(k)}\bigr)\bigr)^2dt \biggr\} \int\exp\biggl\{- \frac{1}{2}\int_0^T \bigl(G^{\alpha }_{t^{(k)}} \bigr)^2dt \biggr\} d\gamma_{\mu} \biggr) \\ &{} \times\biggl( \exp\biggl\{ \int_0^T m_{\mu}^{\alpha}\bigl(t^{(k)}\bigr) dW_t^{\alpha } \biggr\} \\ &{} \times\int\exp\biggl\{ \int_0^T G_{t^{(k)}}^{\alpha} \bigl(dW_t^{\alpha} - m_{\mu}^{\alpha}\bigl(t^{(k)}\bigr)dt \bigr) \biggr\} d \gamma_{\widetilde {K}_{\mu}^{T,k}} \biggr) \biggr\} d\mu \\ = & \log\biggl\{ {\mathcal{E}}_{\mu} \biggl[\exp{ \biggl( - \frac {1}{2} \int_0^T \bigl(G^{\alpha}_{t^{(k)}}\bigr)^2 dt \biggr) } \biggr] \biggr\} - \frac {1}{2}\int_0^T \bigl(m^{\alpha}_{\mu}\bigl(t^{(k)}\bigr) \bigr)^2 dt \\ &{} + \int\int_0^T m_{\mu}^{\alpha}\bigl(t^{(k)}\bigr) dW_t^{\alpha} d\mu \\ &{} +\int\log\biggl\{ \int\exp{ \biggl(\int_0^T G_{t^{(k)}}^{\alpha } \bigl(dW_t^{\alpha} - m_{\mu}^{\alpha}\bigl(t^{(k)}\bigr)dt \bigr) \biggr) } d \gamma_{\widetilde{K}_{\mu}^{T,k}} \biggr\} d\mu, \end{aligned}$$

and standard Gaussian calculus implies that

$$\begin{aligned} &\int\exp{ \biggl(\int_0^T G_{t^{(k)}}^{\alpha} \bigl(dW_t^{\alpha} - m_{\mu }^{\alpha} \bigl(t^{(k)}\bigr)dt \bigr) \biggr) } d\gamma_{\widetilde{K}_{\mu }^{T,k}}\\ &\quad{} = \exp \biggl\{ \frac{1}{2} \int\biggl( \int_0^T G_{t^{(k)}}^{\alpha} \bigl(dW_t^{\alpha}-m_{\mu}^{\alpha} \bigl(t^{(k)}\bigr)dt \bigr) \biggr)^2 d\gamma _{\widetilde {K}_{\mu}^{T,k}} \biggr\}, \end{aligned}$$

so that

$$\begin{aligned} \varGamma^{\alpha, k}(\mu) = &\varGamma^{\alpha, k}_1(\mu) + \int\int_0^T m_{\mu}^{\alpha} \bigl(t^{(k)}\bigr) dW_t^{\alpha} d\mu \\ &{}+ \int\log \biggl\{ \exp\biggl\{ \frac{1}{2} \int\biggl( \int_0^T G_{t^{(k)} }^{\alpha} \bigl(dW_t^{\alpha}-m_{\mu}^{\alpha} \bigl(t^{(k)}\bigr)dt \bigr) \biggr)^2 d\gamma _{\widetilde{K}_{\mu}^{T,k}} \biggr\} \biggr\} d\mu \\ = &\varGamma_1^{\alpha, n}(\mu) + \int\int _0^T m_{\mu}^{\alpha } \bigl(t^{(k)}\bigr) dW_t^{\alpha} d\mu\\ &{}+ \frac{1}{2} \int\int\biggl( \int_0^T G_{t^{(k)} }^{\alpha} \bigl(dW_t^{\alpha}-m_{\mu}^{\alpha} \bigl(t^{(k)}\bigr)dt \bigr) \biggr)^2 d\gamma_{\widetilde{K}_{\mu}^{T,k}} d\mu \end{aligned}$$

which concludes the proof. □

In order to show that H ^k is a good rate function, we need to resort to the definition of two maps defined, for $\mu, \nu\in\mathcal {M}_{1}^{+}(\mathcal{C} )$, by:

$$\begin{aligned} \varGamma^{k}_{\nu}(\mu) = \int_{\mathcal{C}} \log\biggl( \int\exp\biggl\{ \int_0^T \bigl( \mathbf{G}_{t^{(k)}}(\omega)+\mathbf{m}_{\nu}\bigl(t^{(k)} \bigr) \bigr)' \cdot d\mathbf{W} _t(x) \\ - \frac{1}{2} \int_0^T \bigl\| \mathbf{G}_{t^{(k)}}(\omega)+\mathbf{m}_{\nu } \bigl(t^{(k)}\bigr) \bigr\| ^2dt \biggr\} d\gamma_{K_{\nu}}( \omega) \biggr) d\mu(x) \end{aligned}$$

and

$$\begin{aligned} \varGamma_{2,\nu}^{\alpha, k}(\mu) =& \frac{1}{2} \int\int \biggl( \int G^{\alpha}_{t^{(k)}} \bigl(dW^{\alpha}_t - m^{\alpha}_{\nu}\bigl(t^{(k)}\bigr)dt\bigr) \biggr)^2 d\gamma_{\widetilde{K}_{\nu}^{T,n}} d\mu+ \int\int m^{\alpha }_{\nu } \bigl(t^{(k)}\bigr)dW^{\alpha}_t d\mu, \end{aligned}$$

which allows to define $\varGamma_{2,\nu}^{k}$ as the sum over α of $\varGamma_{2,\nu}^{\alpha, k}$ and

$$\begin{aligned} \varGamma_{\nu}^{\alpha,k}=\varGamma_1^{\alpha,k}+ \varGamma_{2,\nu}^{\alpha,k}. \end{aligned}$$

One can easily see that $\varGamma^{k}_{\nu} = \sum_{\alpha=1}^{M} \varGamma _{\nu}^{\alpha, k}$. We eventually introduce a modified H-function:

$$\begin{aligned} H^{k}_{\nu} : \mathcal{M}_1^+(\mathcal{C}) \rightarrow&\mathbb{R}^+ \\ \mu \mapsto&\begin{cases} I(\mu|P) - \varGamma_{\nu}^{k}(\mu) & \mbox{if } I(\mu|P) < \infty,\cr \infty& \mbox{otherwise}. \end{cases} \end{aligned}$$

All these functions enjoy the following properties:

Lemma 5

We recall that d _T denotes the Vaserstein distance (21).

• 1. There exists a positive constant C _T , depending on T but not on n, such that: $|\varGamma_{1}^{k}(\mu)-\varGamma_{1}^{k}(\nu)| \leq C_{T} d_{T}(\mu,\nu)$.

• 2. Γ ^k≤I(.|P) i.e. H ^k is a positive function. In particular, Γ ^k is finite whenever I(.|P) is.

• 3. There exists real constants a<1 and η>0 such that Γ ^k≤aI(.|P)+η.

• 4. There exists a positive constant C _T , depending on T but not on n, such that: $|\varGamma_{2,\nu}^{k}(\mu)-\varGamma_{2}^{k}(\mu)| \leq C_{T} (1+I(\mu|P) ) d_{T}(\mu,\nu)$.

• 5. Defining the following probability measure on $\mathcal {M}_{1}^{+}(\mathcal{C})$:

  $$\begin{aligned} dQ_{\nu}^k(x) = &\exp{\varGamma_{\nu}^{k}( \delta_x)}dP(x) \\ = &\int\exp \biggl( \int_0^T \bigl( \mathbf{G}_{t^{(k)}} + \mathbf{m}_{\nu}\bigl(t^{(k)} \bigr) \bigr)' \cdot\, d\mathbf{W}_t(x)\\ &{} - \frac{1}{2} \int_0^T \bigl\| \mathbf {G}_{t^{(k)}} + \mathbf{m}_{\nu}\bigl(t^{(k)} \bigr) \bigr\|^2 dt \biggr) \, d\gamma_{\nu} \, dP(x) \end{aligned}$$

  we have $H_{\nu}^{k}=I(.|Q_{\nu}^{k})$, so that $H_{\nu}^{k}$ is lower semi-continuous on $\mathcal{M}_{1}^{+}(\mathcal{C})$.

• 6. H ^k is a good rate function.

Once this lemma is proved, it will be easy to demonstrate our Proposition 3 stating that H is a good rate function. In details, we will show that:

Proposition 5

As the partition Δ^k is refined, we have:

• 1. On the compact set $K_{L} = \{\mu\in\mathcal{M}_{1}^{+}(\mathcal {C})| I(\mu|P) \leq L \}$, Γ ^k converges uniformly to Γ.

• 2. $\forall\mu\in K_{L}, \; \varGamma^{\alpha}(\mu) = \varGamma ^{\alpha }_{1}(\mu) + \varGamma^{\alpha}_{2}(\mu)$ where

  

• 3. Γ≤I(.|P) and ∃a<1,η>0| Γ≤aI(.|P)+η.

• 4. H is a good rate function.

Proof

The proof of Lemma 5 (that follows) is easily extended to the continuous case.

(i), (ii):

Similarly to the proof of the assertions (i) and (iv) of Lemma 5, we can find two constants C _1,T and C _2,T such that

$$ \begin{aligned} \bigl| \bigl(\varGamma_1^{\alpha, k}-\varGamma_1^{\alpha, k+p} \bigr) (\mu) \bigr| & \leq C_{1,T} \max_{\gamma=1,M} \biggl( \int\int_0^T \bigl|S_{\alpha \gamma} \bigl(x^{\gamma }_{t^{(k)}-\tau_{\alpha\gamma}}\bigr)\\ &\quad{} - S_{\alpha\gamma} \bigl(x^{\gamma }_{t^{(k+p)}-\tau_{\alpha\gamma}}\bigr) \bigr|^2 dt \, d\mu(x) \biggr)^{\frac{1}{2}} \\ & \leq C_{1,T} \sqrt{T} \max_{\gamma=1,M} \biggl( \int\sup _{|t-s| \leq |\Delta_k|} \bigl|S_{\alpha\gamma}\bigl(x^{\gamma}_t \bigr) - S_{\alpha\gamma }\bigl(x^{\gamma}_s\bigr) \bigr|^2 d\mu(x) \biggr)^{\frac{1}{2}} \\ \bigl| \bigl(\varGamma_2^{\alpha,k}-\varGamma_2^{\alpha, k+p} \bigr) (\mu) \bigr| & \leq C_{2,T} \bigl(I(\mu|P) +1 \bigr) \max _{\gamma=1,M} \biggl( \int\sup_{|t-s| \leq|\Delta_k|} \bigl|S_{\alpha\gamma}\bigl(x^{\gamma}_t\bigr)\\ &\quad{} - S_{\alpha\gamma} \bigl(x^{\gamma }_s\bigr) \bigr|^2 d\mu(x) \biggr)^{\frac{1}{2}}. \end{aligned} $$

But, according to (22), we have for any a≥0

$$\begin{aligned} &a \int\sup_{|t-s| \leq|\Delta_k|} \bigl|S_{\alpha\gamma }\bigl(x^{\gamma}_t \bigr) - S_{\alpha\gamma} \bigl(x^{\gamma}_s\bigr) \bigr|^2 d\mu(x)\\ &\quad{}\leq I(\mu|P) + \log\int\exp{ \Bigl\{ a \sup _{|t-s| \leq|\Delta_k|} \bigl|S_{\alpha\gamma}\bigl(x^{\gamma}_t \bigr) - S_{\alpha\gamma} \bigl(x^{\gamma }_s\bigr) \bigr|^2 \Bigr\}} dP(x). \end{aligned}$$

The bounded convergence theorem ensures that

$$\begin{aligned} \lim_{k\to\infty} \log{\int\exp{ \Bigl\{ a \sup_{|t-s| \leq |\Delta _k|} \bigl|S_{\alpha\gamma}\bigl(x^{\gamma}_t\bigr) - S_{\alpha\gamma } \bigl(x^{\gamma}_s\bigr) \bigr|^2 \Bigr\}} dP(x) } = 0. \end{aligned}$$

Let ε>0, choosing $a=\frac{1}{\varepsilon^{2}}$, it is easy to see that there exists an integer k(ε) such that, for k≥k(ε),∀γ∈{1,…M}:

$$\begin{aligned} \int\sup_{|t-s| \leq|\Delta_k|} \bigl|S_{\alpha\gamma}\bigl(x^{\gamma }_t \bigr) - S_{\alpha\gamma} \bigl(x^{\gamma}_s\bigr) \bigr|^2 d\mu(x) \leq\bigl(I(\mu|P) + 1 \bigr) \varepsilon^2. \end{aligned}$$

Hence, for any k≥k(ε), any p, and any μ∈K _L :

$$\begin{aligned} \bigl| \bigl(\varGamma_1^{\alpha,k}-\varGamma_1^{\alpha,k+p} \bigr) (\mu) \bigr| \leq &C_T (1+L )^{\frac{1}{2}}\varepsilon \\ \bigl| \bigl(\varGamma_2^{\alpha,k}-\varGamma_2^{\alpha,k+p} \bigr) (\mu)\bigr | \leq& C_T (1+L )^{\frac{3}{2}}\varepsilon. \end{aligned}$$

Which shows that $\varGamma_{1}^{\alpha,k}$, $\varGamma_{2}^{\alpha ,k}$, and thus Γ ^α,k converge uniformly on K _L . It is not difficult to see that the respective limits are $\varGamma_{1}^{\alpha}$, $\varGamma _{2}^{\alpha}$ and Γ ^α, which implies $\varGamma ^{\alpha }=\varGamma_{1}^{\alpha}+\varGamma_{2}^{\alpha}$ on K _L . Besides, as $\varGamma^{k} = \sum_{\alpha=1}^{M} \varGamma^{\alpha,k}$ and $\varGamma= \sum_{\alpha=1}^{M} \varGamma^{\alpha}$, we also have the uniform convergence of Γ ^k towards Γ on K _L .

(iii):

This can be proved exactly as that the properties (iii) and (iv) of Lemma 5.

(iv):

We show that {H≤L} is a compact set. H≥(1−a)I(|P)−η so that $I(|P) \leq\frac{H+\eta}{1-a}$. Hence $\{H \leq L \}\subset\{I(|P) \leq\frac{L+\eta }{1-a} \}$. Let $(\mu_{p})_{p} \in\{H \leq L \}^{\mathbb{N}} \subset\{I(|P) \leq\frac{L+\eta}{1-a} \}^{\mathbb{N}}$. As here $\{I(|P) \leq\frac{L+\eta}{1-a} \}$ is a compact set, there exists a subsequence $(\mu_{p_{m}})_{m}$ such that $\mu_{p_{m}} \to \mu$ as m→∞. We conclude by stating that, as H ^k converge uniformly towards H on $\{I(|P) \leq\frac{L+\eta}{1-a} \}$, the latest inherits the lower semi-continuity of the firsts. Hence {H≤L} is a closed set so that μ∈{H≤L} and $(\mu_{p_{m}})_{m}$ converges in {H≤L}.

 □

We are hence left proving Lemma 5.

Proof of Lemma 5.(i)

We prove Lipschitz-continuity for $\varGamma^{\alpha, k}_{1}$. We have:

$$\begin{aligned} &\biggl|\log\biggl( 1 + \frac{\int\exp( - \frac{1}{2} \sum_{l=0}^k {G^{\alpha}_{t_l}}^2(t_{l+1} - t_l) ) d(\gamma_{K_{\mu}}-\gamma _{K_{\nu}})}{\int\exp( - \frac{1}{2} \sum_{l=0}^k {G^{\alpha }_{t_l}}^2(t_{l+1} - t_l) ) d\gamma_{K_{\nu}}} \biggr) \biggr| \\ &\quad{}= \biggl|\varGamma _1^{\alpha,k}( \mu)-\varGamma_1^{\alpha,k}(\nu)+\frac {1}{2}\int\bigl( \bigl(m^{\alpha}_{\mu}\bigr)^2-\bigl(m^{\alpha}_{\nu} \bigr)^2\bigr) \bigl(t^{(k)}\bigr)dt \biggr| \\ &\quad{} \leq\exp\biggl\{\frac{k_{\alpha}T}{2\lambda_{\alpha}^2} \biggr\} \biggl|\int\exp\biggl( - \frac {1}{2} \int_0^T {G^{\alpha}_{t^{(k)}}}^2dt \biggr) d(\gamma_{K_{\mu }}-\gamma_{K_{\nu}}) \biggr|. \end{aligned}$$

Let ξ be a probability measure on $\mathcal{C}\times\mathcal {C}$ with marginals μ and ν, and let γ _ξ be the law of a bidimensional centered Gaussian process $(G^{\alpha}, \widetilde{G^{\alpha}})$ with covariance $K^{\alpha}_{\xi}$:

$$ \begin{aligned}[b] &K^{\alpha}_{\xi}(s,t)\\ &\quad{} = \sum_{\gamma=1}^M \frac{\sigma_{\alpha \gamma}^2}{\lambda_{\alpha}^2} \left ( \begin{array}{c@{\quad}c} \int S_{\alpha\gamma}(x^{\gamma}_{s-\tau_{\alpha\gamma }})S_{\alpha\gamma}(x^{\gamma}_{t-\tau_{\alpha\gamma}}) \, d\xi(x,y) & \int S_{\alpha\gamma}(x^{\gamma}_{s-\tau_{\alpha\gamma }})S_{\alpha\gamma}(y^{\gamma}_{t-\tau_{\alpha\gamma}}) \, d\xi (x,y) \\ \int S_{\alpha\gamma}(y^{\gamma}_{s-\tau_{\alpha\gamma }})S_{\alpha\gamma}(x^{\gamma}_{t-\tau_{\alpha\gamma}}) \, d\xi(x,y) & \int S_{\alpha\gamma}(y^{\gamma}_{s-\tau_{\alpha\gamma }})S_{\alpha\gamma}(y^{\gamma}_{t-\tau_{\alpha\gamma}}) \, d\xi (x,y) \\ \end{array} \right ) . \end{aligned} $$

(34)

Then,

$$\begin{aligned} &\biggl|\int\exp\biggl( - \frac{1}{2} \int_0^T {G^{\alpha}_{t^{(k)} }}^2dt \biggr) d( \gamma_{K_{\mu}}-\gamma_{K_{\nu}}) \biggr|\\ &\quad{} = \biggl|\int\biggl\{ \exp\biggl( - \frac{1}{2} \int_0^T {G^{\alpha}_{t^{(k)}}}^2dt \biggr) - \exp\biggl( - \frac {1}{2} \int_0^T \widetilde{G^{\alpha}_{t^{(k)}}}^2dt \biggr) \biggr\} d \gamma_{\xi} \biggr| \\ &\quad{} \leq\frac{1}{2} \int\int_0^T \bigl| {G^{\alpha}_{t^{(k)} }}^2-\widetilde{G^{\alpha}_{t^{(k)}}}^2 \bigr| dt d\gamma_{\xi} \\ &\quad{} \leq\frac{1}{2} \prod_{\varepsilon= \pm1} \biggl(\int\int _0^T \bigl(G^{\alpha }_{t^{(k)}}+ \varepsilon\widetilde{G^{\alpha}_{t^{(k)}}}\bigr)^2 dt d \gamma_{\xi} \biggr)^{\frac{1}{2}} \end{aligned}$$

by Cauchy-Schwarz inequality. Then, using the covariance of $(G^{\alpha },\widetilde{G^{\alpha}})$ under γ _ξ , we find:

$$\begin{aligned} & \biggl|\varGamma_1^{\alpha,k}(\mu)-\varGamma_1^{\alpha,k}( \nu)+\frac{1}{2}\int\bigl(\bigl(m^{\alpha}_{\mu} \bigr)^2-\bigl(m^{\alpha}_{\nu}\bigr)^2 \bigr) \bigl(t^{(k)}\bigr)dt \biggr| \\ & \quad{}\leq\frac{1}{2} \exp\biggl\{\frac{k_{\alpha}T}{2\lambda _{\alpha}^2} \biggr\} \biggl( \frac{4k_{\alpha}T}{\lambda_{\alpha}^2} \biggr)^{\frac{1}{2}} \Biggl\{ \frac{1}{\lambda_{\alpha} ^2}\sum _{\gamma=1}^M \sigma_{\alpha\gamma}^2 \int\int_0^T \bigl(S_{\alpha \gamma} \bigl(x^{\gamma}_{t-\tau_{\alpha\gamma}}\bigr) \\ & \qquad{}-S_{\alpha\gamma} \bigl(y^{\gamma}_{t-\tau_{\alpha\gamma}}\bigr)\bigr)^2 dt \, d\xi(x,y) \Biggr\} ^{\frac{1}{2}} \\ & \quad{}\leq\frac{k_{\alpha}}{\lambda_{\alpha}^2} \sqrt{T} \exp\biggl\{ \frac{k_{\alpha} T}{2\lambda_{\alpha} ^2} \biggr \} \max_{\gamma=1\cdots M} \biggl\{ \int\int_0^T \bigl|S_{\alpha\gamma} \bigl(x^{\gamma}_{t-\tau_{\alpha\gamma}}\bigr )-S_{\alpha\gamma} \bigl(y^{\gamma }_{t-\tau_{\alpha\gamma}}\bigr) \bigr|^2 dt \, d\xi(x,y) \biggr\}^{\frac{1}{2}}. \end{aligned}$$

(35)

Moreover, we have:

$$\begin{aligned} \biggl|\int{m^{\alpha}_{\mu}\bigl(t^{(k)} \bigr)}^2 -{m^{\alpha}_{\nu}\bigl(t^{(k)} \bigr)}^2 dt \biggr| = &\int\bigl| \bigl(m^{\alpha}_{\mu} \bigl(t^{(k)}\bigr) -m^{\alpha}_{\nu} \bigl(t^{(k)} \bigr) \bigr) \bigl(m^{\alpha}_{\mu} \bigl(t^{(k)}\bigr) + m^{\alpha}_{\nu} \bigl(t^{(k)}\bigr) \bigr) \bigr| dt \\ \leq&2 \frac{\bar{J}_{\alpha}}{\lambda_{\alpha}} \int\bigl|m^{\alpha}_{\mu} \bigl(t^{(k)}\bigr) -m^{\alpha }_{\nu} \bigl(t^{(k)}\bigr) \bigr| dt. \end{aligned}$$

But

$$\begin{aligned} \int_0^T \bigl|\bigl(m^{\alpha}_{\mu}-m^{\alpha}_{\nu} \bigr) (t)\bigr| dt = &\int_0^T \Biggl|\frac{1}{\lambda_{\alpha}} \sum_{\gamma=1}^M \bar{J}_{\alpha \gamma}\int S_{\alpha\gamma}\bigl(x^{\gamma }_{t-{\tau}}\bigr) d(\mu-\nu) (x) \Biggr| \, dt \\ \leq&\frac{1}{\lambda_{\alpha}} \sum_{\gamma=1}^M \bigl| \bar{J}_{\alpha\gamma}\bigr| \int_0^T \biggl|\int S_{\alpha\gamma}\bigl(x^{\gamma}_{t-{\tau_{\alpha\gamma}}}\bigr) d(\mu -\nu) (x) \biggr| \, dt \\ \leq&\frac{1}{\lambda_{\alpha}} \sum_{\gamma=1}^M | \bar{J}_{\alpha\gamma}| \int\int_0^T \bigl|S_{\alpha\gamma} \bigl(x^{\gamma}_{t-{\tau_{\alpha\gamma}}}\bigr) - S_{\alpha\gamma }\bigl(y^{\gamma}_{t-{\tau}}\bigr)\bigr| dt \, d\xi(x,y) \\ \leq&\frac{ \bar{J}_{\alpha}}{\lambda_{\alpha}} \max_{\gamma =1\cdots M} \biggl( \int\int _0^T \bigl|S_{\alpha\gamma}\bigl(x^{\gamma}_{t-{\tau_{\alpha\gamma}}} \bigr) - S_{\alpha\gamma}\bigl(y^{\gamma}_{t-{\tau }}\bigr) \bigr|^2 dt \, d\xi(x,y) \biggr)^{\frac{1}{2}} \end{aligned}$$

by Cauchy-Schwarz inequality.

Consequently,

$$\begin{aligned} \bigl|\varGamma_1^{\alpha,k}(\mu) - \varGamma_1^{\alpha,k}( \nu)\bigr| \leq&\biggl(\frac{ \bar{J}_{\alpha}^2}{\lambda_{\alpha}^2} + \frac{k_{\alpha }}{\lambda_{\alpha}^2} \sqrt{T} \exp\biggl\{ \frac{k_{\alpha} T}{2\lambda_{\alpha} ^2} \biggr\} \biggr) \max_{\gamma=1\cdots M} \biggl( \int \int_0^T \bigl|S_{\alpha\gamma} \bigl(x^{\gamma}_{t^{(k)}-\tau_{\alpha\gamma}} \bigr) \\ &{} - S_{\alpha\gamma }\bigl(y^{\gamma}_{t^{(k)}-\tau_{\alpha\gamma}}\bigr) \bigr|^2 dt \, d\xi(x,y) \biggr)^{\frac{1}{2}} . \end{aligned}$$

(36)

As the S _αγ are K _S Lipschitz, we have:

$$\begin{aligned} \bigl|\varGamma_1^{\alpha,k}(\mu) - \varGamma_1^{\alpha,k}( \nu)\bigr| \leq K_S \sqrt{T} \biggl(\frac{\bar{J}_{\alpha}^2}{\lambda _{\alpha}^2} + \frac {k_{\alpha}}{\lambda_{\alpha}^2} \sqrt{T} \exp\biggl\{ \frac{k_{\alpha }T}{2\lambda_{\alpha}^2} \biggr\} \biggr) d_T(\mu,\nu), \end{aligned}$$

(37)

so that $\varGamma_{1}^{\alpha,k}$ is Lipschitz for the Vaserstein distance. Using the triangle inequality, the result holds for $\varGamma_{1}^{k}$.

Proof of Lemma 5.(ii)

Let

$$\begin{aligned} F_{\mu}(x) =& \log\biggl\{ \int\exp\biggl\{ \int_0^T \bigl(\mathbf{G}_{t^{(k)} }(\omega)+\mathbf{m}_{\mu} \bigl(t^{(k)}\bigr) \bigr)' \cdot d\mathbf{W}_t(x) \\ &{}- \frac{1}{2} \int_0^T \bigl\| \mathbf{G}_{t^{(k)}}(\omega)+\mathbf{m}_{\mu } \bigl(t^{(k)}\bigr) \bigr\|^2 dt \biggr\} d\gamma_{\mu} \biggr\}. \end{aligned}$$

This function is a.s. finite but not bounded, let us hence define for $A \in\mathbb{R}^{+}$

$$\begin{aligned} F_{\mu}^A(x) =& \log\biggl\{ \int A\wedge\exp\biggl\{ \int_0^T \bigl(\mathbf{G}_{t^{(k)} }( \omega)+\mathbf{m}_{\mu}\bigl(t^{(k)}\bigr) \bigr)' \cdot d\mathbf{W}_t(x) \\ &{}- \frac{1}{2} \int_0^T \bigl\| \mathbf{G}_{t^{(k)}}(\omega)+\mathbf{m}_{\mu } \bigl(t^{(k)}\bigr) \bigr\|^2 dt \biggr\} d\gamma_{\mu} \biggr\}. \end{aligned}$$

By the monotone convergence theorem and using Eq. (22), we have for any a≥1:

$$\begin{aligned} a \int F_{\mu}(x) d\mu(x) \leq I(\mu|P)+ \log\biggl\{ \int\exp{ aF_{\mu }(x)} dP(x) \biggr\}. \end{aligned}$$

By Jensen inequality and Fubini theorem,

$$\begin{aligned} \int\exp{ \bigl(aF_{\mu}(x)}\bigr) dP(x) \leq&\int\prod _{\alpha=1}^M \int\exp\biggl\{ a \int _0^T \bigl(G^{\alpha}_{t^{(k)}} + m^{\alpha}_{\mu }\bigl(t^{(k)}\bigr) \bigr) dW^{\alpha}_t(x) \biggr\} dP_{\alpha}(x) \\ &{}\times \exp\biggl\{ - \frac{a}{2} \int_0^T \bigl\|\mathbf{G}_{t^{(k)}}(\omega)+\mathbf{m}_{\mu} \bigl(t^{(k)} \bigr) \bigr\|^2 dt \biggr\} d\gamma_{\mu}. \end{aligned}$$

But, as W ^α is a P _α -Brownian motion,

$$\begin{aligned} \int\exp\biggl\{ a \int_0^T \bigl(G^{\alpha}_{t^{(k)}} + m^{\alpha }_{\mu} \bigl(t^{(k)} \bigr) \bigr) dW^{\alpha}_t(x) \biggr\} dP_{\alpha}(x) = \exp\biggl\{ \frac {a^2}{2} \int _0^T \bigl(G^{\alpha}_{t^{(k)}} + m^{\alpha}_{\mu}\bigl(t^{(k)}\bigr) \bigr)^2 dt \biggr\} , \end{aligned}$$

so that

$$\begin{aligned} a \int F_{\mu}(x) d\mu(x) \leq I(\mu|P)+ \log\biggl\{ \int\exp\biggl \{ \frac{a^2-a}{2} \int_0^T \bigl\| \mathbf{G}_{t^{(k)}}(\omega)+\mathbf{m}_{\mu } \bigl(t^{(k)}\bigr)\bigr \|^2 dt \biggr\} d\gamma_{\mu} \biggr\}. \end{aligned}$$

Letting a=1 proves that Γ ^k≤I(|P).

Proof of Lemma 5.(iii)

As the components of G are independent under γ _μ , we only have to check that, for every b>0, there exists a finite constant C _b such that

$$\begin{aligned} {\mathcal{E}}_{\mu} \biggl[ \exp{ \biggl(\frac{b}{2} \int _0^T \bigl(G^{\alpha }_s+m^{\alpha}_{\mu}(s) \bigr)^2 ds \biggr) } \biggr] \leq\exp{\frac {b C_b k_{\alpha}T}{\lambda_{\alpha}^2}} . \end{aligned}$$

(38)

It was proved in [3, Lemma A.3(2)] in their particular framework that for every b verifying $\frac{bk_{\alpha }T}{\lambda_{\alpha} ^{2}}<1$, there exists a finite constant c _b such that:

$$\begin{aligned} \int\exp{ \biggl(\frac{b}{2} \int_0^T G_s^2 ds \biggr) } d\gamma_{\mu } \leq\exp{ \frac{b c_b k_{\alpha}T}{\lambda_{\alpha}^2}}. \end{aligned}$$

In our case, the covariance function is slightly different of that of [3], but the proof and result remain unchanged and can be readily extended.

Moreover, since we have

$$\begin{aligned} \bigl(G^{\alpha}_s + m^{\alpha}_{\mu}(s) \bigr)^2 \leq2 {G^{\alpha}_s}^2 + 2 {m^{\alpha}_{\mu}(s)}^2 \leq2{G^{\alpha}_s}^2 + 2\frac{\bar {J}_{\alpha} ^2}{\lambda_{\alpha}^2} \end{aligned}$$

we obtain the desired result with the following constant $C_{b} = 2 c_{2b} + \frac{\bar{J}_{\alpha}^{2}}{k_{\alpha}}$, under the condition $\frac{2b k_{\alpha} T}{\lambda_{\alpha} ^{2}}<1$.

Proof of Lemma 5.(iv)

As above, let us prove the result for $| \varGamma^{\alpha ,k}_{2,\nu} - \varGamma^{\alpha,k}_{2} |$. We have:

$$\begin{aligned} \bigl|\varGamma_{2,\nu}^{\alpha,k}(\mu) - \varGamma_2^{\alpha,k}( \mu)\bigr| \leq&\frac {1}{2} \biggl|\int\int\biggl\{ \biggl( \int G^{\alpha}_{t^{(k)}} \bigl(dW^{\alpha }_t - m^{\alpha}_{\mu} \bigl(t^{(k)}\bigr)dt\bigr) \biggr)^2 \\ &{}- \biggl( \int G^{\alpha }_{t^{(k)}} \bigl(dW^{\alpha}_t - m^{\alpha}_{\nu}\bigl(t^{(k)}\bigr)dt\bigr) \biggr)^2 \biggr\} d\gamma_{\widetilde{K}_{\mu}^{T,k}} d\mu\biggr| \\ &{}+\frac{1}{2} \biggl|\int\int\biggl( \int G^{\alpha}_{t^{(k)}} \bigl(dW^{\alpha}_t - m^{\alpha}_{\nu } \bigl(t^{(k)}\bigr)dt\bigr) \biggr)^2 d ( \gamma_{\widetilde{K}_{\nu}^{T,k}}-\gamma_{\widetilde {K}_{\mu}^{T,k}} ) d\mu\biggr|\\ &{} + \biggl|\int\int \bigl(m^{\alpha}_{\nu }-m^{\alpha }_{\mu}\bigr) \bigl(t^{(k)}\bigr)dW^{\alpha}_t d\mu\biggr|. \end{aligned}$$

Let ξ be a probability measure on $\mathcal{C}\times\mathcal {C}$ with marginals μ and ν, and let γ _ξ be the law of a bidimensional centered Gaussian process $(G^{\alpha}, \widetilde{G^{\alpha}})$ with covariance $K^{\alpha}_{\xi}$. Let

$$\begin{aligned} \varLambda_T^{\alpha,k}\bigl(G^{\alpha}\bigr)= \frac{\exp{ ( -\frac {1}{2} \int_0^T {G^{\alpha}_{t^{(k)}}}^2 dt )}}{\int\exp{ ( -\frac {1}{2} \int_0^T {G^{\alpha}_{t^{(k)}}}^2 dt )} \, d\gamma_{\xi}}. \end{aligned}$$

As in [3, Lemma 3.4], we can show that:

$$ \begin{aligned}[b] &\bigl|\varGamma_{2,\nu}^{\alpha,k}(\mu) - \varGamma_2^{\alpha,k}( \mu)\bigr|\\ &\quad{} \leq\frac {1}{2} \overbrace{\int\int\bigl|\varLambda_T^{\alpha,k} \bigl(G^{\alpha }\bigr)-\varLambda_T^{\alpha,k}\bigl( \widetilde{G^{\alpha}}\bigr) \bigr| \biggl( \int G^{\alpha}_{t^{(k)}} \bigl(dW^{\alpha}_t - m^{\alpha}_{\nu} \bigl(t^{(k)}\bigr)dt\bigr) \biggr)^2 d\gamma_{\xi} d\mu}^{B_1} \\ &\qquad{}+ \underbrace{\frac{1}{2} \! \prod_{\varepsilon=\pm1} \biggl( \int\! \int\varLambda_T^{\alpha ,k}\bigl(\widetilde {G^{\alpha}}\bigr) \biggl( \int\bigl(G^{\alpha}_{t^{(k)}} + \varepsilon\widetilde{G^{\alpha}}_{t^{(k)}}\bigr) \bigl(dW^{\alpha}_t - m^{\alpha}_{\nu} \bigl(t^{(k)}\bigr)dt\bigr) \biggr)^2 d\gamma_{\xi} d\mu\biggr)^{\frac{1}{2}}}_{B_2} \\ &\qquad{}+ \frac{1}{2} \underbrace{ \biggl|\int\!\!\int\varLambda_T^{\alpha,k} \bigl(G^{\alpha}\bigr) \biggl\{ \biggl( \int G^{\alpha}_{t^{(k)}} \bigl(dW^{\alpha}_t - m^{\alpha}_{\mu} \bigl(t^{(k)}\bigr)dt\bigr) \biggr)^2 - \biggl( \int G^{\alpha}_{t^{(k)}} \bigl(dW^{\alpha}_t - m^{\alpha}_{\nu }\bigl(t^{(k)}\bigr)dt\bigr) \biggr)^2 \biggr\} d\gamma_{\xi} d\mu\biggr|}_{B_3} \\ &\qquad{}+ \underbrace{ \biggl( \int\biggl| \int\bigl(m^{\alpha}_{\nu}-m^{\alpha}_{\mu} \bigr) \bigl(t^{(k)} \bigr)dW^{\alpha}_t \biggr|^2 d\mu\biggr)^{\frac{1}{2}}}_{B_4} \end{aligned} $$

(39)

and

$$\begin{aligned} \varLambda_T^{\alpha,k}\bigl(G^{\alpha}\bigr)= \exp\biggl \{ - \varGamma_1^{\alpha,k}(\mu) - \frac{1}{2} \int _0^T {m^{\alpha}_{\mu}}^2 \bigl(t^{(k)}\bigr) dt - \frac {1}{2} \int_0^T {G^{\alpha}_{t^{(k)}}}^2 dt \biggr\}. \end{aligned}$$

Hence, we have by Jensen inequality,

$$\begin{aligned} \varLambda_T^{\alpha,k}\bigl(G^{\alpha}\bigr) \leq\exp \biggl\{\frac {k_{\alpha} T}{2\lambda_{\alpha}^2} \biggr\}, \end{aligned}$$

so that

$$\begin{aligned} \bigl|\varLambda_T^{\alpha,k}\bigl(G^{\alpha}\bigr)- \varLambda_T^{\alpha ,k}\bigl(\widetilde{G^{\alpha}}\bigr) \bigr| \leq&\exp\biggl\{\frac{k_{\alpha}T}{2\lambda _{\alpha}^2} \biggr\} \biggl(\frac {1}{2}\int _0^T \bigl| {G^{\alpha}_{t^{(k)}}}^2- \widetilde{G^{\alpha }}_{t^{(k)} }^2 \bigr|dt\\ &{} + \biggl| \varGamma_1^{\alpha,k}(\mu)-\varGamma_1^{\alpha ,k}( \nu)+\frac{1}{2}\int\bigl(\bigl(m^{\alpha}_{\mu} \bigr)^2-\bigl(m^{\alpha}_{\nu}\bigr)^2 \bigr) \bigl(t^{(k)} \bigr)dt \biggr| \biggr), \end{aligned}$$

which eventually gives

$$\begin{aligned} B_1 \leq&\frac{1}{2}\exp\biggl\{\frac{k_{\alpha}T}{2\lambda_{\alpha }^2} \biggr\} \biggl( \biggl|\varGamma_1^{\alpha,k}(\mu)-\varGamma_1^{\alpha,k}( \nu)+\frac {1}{2}\int\bigl(\bigl(m^{\alpha}_{\mu} \bigr)^2-\bigl(m^{\alpha}_{\nu}\bigr)^2 \bigr) \bigl(t^{(k)}\bigr)dt \biggr|\\ &{}\times \int\int\biggl( \int _0^T G^{\alpha}_{t^{(k)}} \bigl(dW^{\alpha}_t - m^{\alpha}_{\nu} \bigl(t^{(k)} \bigr)dt\bigr) \biggr)^2 d\gamma_{\xi}d \mu \\ &{}+ \int\int\biggl(\int_0^T \bigl| \bigl(G^{\alpha}_{t^{(k)}}\bigr)^2-\widetilde {G^{\alpha }}_{t^{(k)}}^2 \bigr| dt \biggr) \biggl( \int _0^T G^{\alpha}_{t^{(k)}} \bigl(dW^{\alpha}_t - m^{\alpha}_{\nu} \bigl(t^{(k)}\bigr)dt\bigr) \biggr)^2 d\gamma_{\xi}d \mu\biggr). \end{aligned}$$

Let h,m∈L ²([0;T],dt), with m bounded. By Cauchy-Schwarz and the relative entropy inequality (22) (with $\varPhi(x)= ( \int_{0}^{T} h_{t} dW^{\alpha}_{t}(x) )^{2} \sim \mathcal {N} (0,\int_{0}^{T} h_{t}^{2} dt )^{2}$ under P _α , and besides is a positive and measurable function of $\mathcal{C}$), we have the existence of a finite constant C such that,

$$\begin{aligned} \int\biggl( \int_0^T h_t \bigl(dW^{\alpha}_t(x) - m(t)dt\bigr) \biggr)^2 d\mu (x) \leq&2 \biggl\{ \int\biggl(\int_0^T h_t dW^{\alpha}_t \biggr)^2 + \biggl( \int_0^T h_t m_t dt \biggr)^2 d\mu\biggr\} \\ \leq&2\biggl \{ \bigl(C \bigl(1+I(\mu|P) \bigr) + m^2_{\infty}T \bigr) \biggl( \int_0^T h_t^2 dt \biggr)\biggr \} \\ \leq &C' \bigl(1+I(\mu|P) \bigr) \biggl( \int _0^T h_t^2 dt \biggr) . \end{aligned}$$

(40)

We can now bound the different terms in inequality (39).

In fact, as $h_{t}=G^{\alpha}_{t^{(k)}}$ and $m_{t}=m^{\alpha}_{\nu}(t^{(k)})$ verify the required condition, (40) gives the existence of c _T ,

$$\begin{aligned} \int\biggl( \int_0^T G^{\alpha}_{t^{(k)}} \bigl(dW^{\alpha}_t(x) - m^{\alpha }_{\nu } \bigl(t^{(k)}\bigr)dt\bigr) \biggr)^2 d\mu(x) \leq\: c_T \bigl(1+I(\mu|P) \bigr) \int_0^T \bigl(G^{\alpha}_{t^{(k)}}\bigr)^2 dt. \end{aligned}$$

Hence, we can find a finite constant $c'_{T}$ such that

$$\begin{aligned} B_1 \leq c'_T \bigl(1+I(\mu|P) \bigr) \max_{\gamma=1\cdots M} \biggl( \int\int_0^T \bigl|S_{\alpha\gamma}\bigl(x^{\gamma}_{t^{(k)}-\tau_{\alpha \gamma}}\bigr)-S_{\alpha\gamma} \bigl(y^{\gamma}_{t^{(k)} -\tau_{\alpha\gamma} }\bigr) \bigr|^2 dt \, d\xi(x,y) \biggr)^{\frac{1}{2}}. \end{aligned}$$

Similarly, there exists a constant c _T such that

$$\begin{aligned} B_2 \leq&\frac{1}{2}\exp\biggl\{\frac{k_{\alpha}T}{2\lambda _{\alpha}^2} \biggr \} \prod_{\varepsilon=\pm1} \biggl( c_T \bigl(1+I( \mu|P) \bigr) \int\int\bigl(G^{\alpha }_{t^{(k)}} +\varepsilon \widetilde{G^{\alpha}}_{t^{(k)}}\bigr)^2 dt d\gamma _{\xi} \biggr)^{\frac{1}{2}} \\ \leq &c'_T \bigl(1+I(\mu|P) \bigr) \max _{\gamma=1\cdots M} \biggl( \int\int_0^T \bigl|S_{\alpha\gamma}\bigl(x^{\gamma}_{t^{(k)}-\tau_{\alpha\gamma }}\bigr)-S_{\alpha\gamma} \bigl(y^{\gamma}_{t^{(k)} -\tau_{\alpha\gamma}}\bigr) \bigr|^2 dt \, d\xi(x,y) \biggr)^{\frac{1}{2}}. \end{aligned}$$

To bound B ₃, we first use Cauchy-Schwarz inequality:

$$\begin{aligned} B_3 \leq&\frac{1}{2} \exp\biggl\{ \frac{k_{\alpha}T}{2\lambda _{\alpha}^2} \biggr\} \prod_{\varepsilon= \pm1} \biggl\{ \int\int\biggl| \int _0^T G^{\alpha }_{t^{(k)} } \bigl( (1+ \varepsilon)dW^{\alpha}_t \\ &{}- \bigl(m^{\alpha}_{\nu} \bigl(t^{(k)}\bigr) + \varepsilon m^{\alpha}_{\mu} \bigl(t^{(k)}\bigr)\bigr)dt \bigr)\biggr |^2 d\gamma _{\xi} d\mu\biggr\}^{\frac{1}{2}} . \end{aligned}$$

(41)

But

$$\begin{aligned} \biggl| \int_0^T G^{\alpha}_{t^{(k)}} \bigl(m^{\alpha}_{\mu}\bigl(t^{(k)}\bigr) -m^{\alpha }_{\nu}\bigl(t^{(k)}\bigr) \bigr) dt \biggr|^2 \leq\biggl(\int_0^T {G_{t^{(k)} }^{\alpha}}^2 dt \biggr) \biggl( \int _0^T \bigl(m^{\alpha}_{\mu} \bigl(t^{(k)}\bigr) -m^{\alpha }_{\nu} \bigl(t^{(k)} \bigr) \bigr)^2 dt \biggr). \end{aligned}$$

Remark that

$$\begin{aligned} \int_0^T \bigl(m^{\alpha}_{\mu} \bigl(t^{(k)}\bigr) - m^{\alpha}_{\nu } \bigl(t^{(k)}\bigr) \bigr)^2 dt = &\int _{0}^T \Biggl( \sum _{\gamma=1}^M \frac{\bar {J}_{\alpha\gamma}}{\lambda_{\alpha}} \int S_{\alpha\gamma} \bigl(x^{\gamma}_{t^{(k)}-\tau_{\alpha\gamma}}\bigr) d(\mu- \nu) (x) \Biggr)^2 dt \\ \leq&\frac{\bar{J}_{\alpha}^2}{\lambda_{\alpha}^2} \max_{\gamma =1\cdots M} \int _{0}^T \biggl( \int S_{\alpha\gamma} \bigl(x^{\gamma}_{t^{(k)}-\tau_{\alpha\gamma}}\bigr) d(\mu- \nu) (x) \biggr)^2 dt \\ \leq&\frac{\bar{J}_{\alpha}^2}{\lambda_{\alpha}^2} \max_{\gamma =1\cdots M} \int _{0}^T \biggl( \int\bigl|S_{\alpha\gamma} \bigl(x^{\gamma}_{t^{(k)}-\tau_{\alpha\gamma}}\bigr)\\ &{} - S_{\alpha\gamma} \bigl(y^{\gamma}_{t^{(k)}-\tau_{\alpha\gamma} }\bigr)\bigr| d\xi(x,y) \biggr)^2 dt. \end{aligned}$$

So that

$$\begin{aligned} &\biggl| \int_0^T G^{\alpha}_{t^{(k)}} \bigl(m^{\alpha}_{\mu}\bigl(t^{(k)}\bigr) -m^{\alpha }_{\nu}\bigl(t^{(k)}\bigr) \bigr) dt \biggr|^2 \\ &\quad{}\leq\frac{\bar{J}_{\alpha}^2}{\lambda_{\alpha}^2} \biggl(\int_0^T G^2_{t^{(k)}} dt \biggr) \max_{\gamma =1\cdots M} \int _{0}^T \biggl( \int\bigl|S_{\alpha\gamma} \bigl(x^{\gamma }_{t^{(k)}-\tau_{\alpha\gamma}}\bigr) - S_{\alpha\gamma} \bigl(y^{\gamma}_{t^{(k)}-\tau_{\alpha\gamma}}\bigr)\bigr| d\xi(x,y) \biggr )^2 dt. \end{aligned}$$

Moreover, (40) gives:

$$\begin{aligned} \int\biggl\{ \int_0^T 2 G^{\alpha}_{t^{(k)}} \biggl(dW^{\alpha}_t - \frac {m^{\alpha}_{\mu}(t^{(k)})+m^{\alpha}_{\nu}(t^{(k)})}{2} dt \biggr) \biggr\} ^2 d\mu \leq c_T \bigl( 1+ I(\mu|P) \bigr) 4 \int _0^T {G^{\alpha}_{t^{(k)} }}^2 dt. \end{aligned}$$

Using the last two inequalities in (41) we have:

$$\begin{aligned} B_3 \leq&\frac{1}{2} \exp\biggl\{ \frac{k_{\alpha}T}{2\lambda _{\alpha}^2} \biggr\} \biggl\{ \int c_T \bigl( 1+ I(\mu|P) \bigr) 4 \biggl(\int _0^T {G^{\alpha}_{t^{(k)}}}^2 dt \biggr) d\gamma_{\xi} \biggr\}^{\frac{1}{2}} \\ &{}\times\biggl\{ \int \frac{\bar{J}_{\alpha}^2}{\lambda_{\alpha}^2} \biggl(\int_0^T {G^{\alpha}_{t^{(k)}}}^2 dt \biggr) \max _{\gamma=1\cdots M} \int_{0}^T \biggl( \int\bigl|S_{\alpha \gamma} \bigl(x^{\gamma }_{t^{(k)}-\tau_{\alpha\gamma}}\bigr)\\ &{} -S_{\alpha\gamma}\bigl(y^{\gamma }_{t^{(k)}-\tau_{\alpha\gamma}}\bigr)\bigr| d\xi(x,y) \biggr)^2 dt d\gamma_{\xi} \biggr\}^{\frac{1}{2}} \\ \leq &c'_T \bigl( 1+ I(\mu|P) \bigr) \max _{\gamma=1\cdots M} \biggl\{\int_{0}^T \int\bigl|S_{\alpha\gamma}\bigl(x^{\gamma}_{t^{(k)}-\tau_{\alpha \gamma}}\bigr) - S_{\alpha\gamma}\bigl(y^{\gamma}_{t^{(k)} -\tau_{\alpha\gamma} }\bigr)\bigr|^2 d \xi(x,y) dt \biggr\}^{\frac{1}{2}} \end{aligned}$$

as I(|P)≥0.

As of the last term, we have

$$\begin{aligned} B_4 \leq&\biggl( c_T \bigl(1+I(\mu|P) \bigr) \int _0^T \bigl(m^{\alpha}_{\mu } \bigl(t^{(k)}\bigr) - m^{\alpha}_{\nu} \bigl(t^{(k)}\bigr) \bigr)^2 dt \biggr)^{\frac {1}{2}} \\ \leq&\frac{\bar{J}_{\alpha}}{\lambda_{\alpha}} \bigl( c_T \bigl(1+I(\mu |P) \bigr) \bigr)^{\frac {1}{2}} \max_{\gamma=1\cdots M} \biggl(\int _{0}^T \biggl( \int\bigl|S_{\alpha\gamma} \bigl(x^{\gamma}_{t^{(k)}-\tau_{\alpha\gamma}}\bigr)\\ &{} - S_{\alpha\gamma } \bigl(y^{\gamma}_{t^{(k)}-\tau_{\alpha\gamma}}\bigr)\bigr| d\xi(x,y) \biggr )^2 dt \biggr)^{\frac{1}{2}} \\ \leq &c'_T \bigl(1+I(\mu|P) \bigr) \max _{\gamma=1\cdots M} \biggl(\int_{0}^T \int\bigl|S_{\alpha\gamma}\bigl(x^{\gamma}_{t^{(k)}-\tau_{\alpha \gamma}}\bigr) - S_{\alpha\gamma}\bigl(y^{\gamma}_{t^{(k)} -\tau_{\alpha\gamma} }\bigr)\bigr|^2 d \xi(x,y) dt \biggr)^{\frac{1}{2}}. \end{aligned}$$

We have proved that there exist a constant c _T such that

$$ \begin{aligned}[b] &\bigl|\varGamma_{2,\nu}^{\alpha,k}(\mu) - \varGamma_2^{\alpha,k}( \mu)\bigr|\\ &\quad{}\leq c_T \bigl(1+I(\mu|P) \bigr) \max_{\gamma=1\cdots M} \biggl(\int_{0}^T \int\bigl|S_{\alpha\gamma} \bigl(x^{\gamma}_{t^{(k)}-\tau_{\alpha\gamma}}\bigr) - S_{\alpha\gamma } \bigl(y^{\gamma}_{t^{(k)}-\tau_{\alpha\gamma}}\bigr)\bigr|^2 d\xi(x,y) dt \biggr)^{\frac{1}{2}}. \end{aligned} $$

(42)

Therefore

$$\begin{aligned} \bigl|\varGamma_{2,\nu}^{\alpha,k}(\mu) - \varGamma_2^{\alpha,k}( \mu)\bigr| \leq c_T K_S \sqrt{T} \bigl(1+I(\mu|P) \bigr) d_T(\mu,\nu), \end{aligned}$$

so that using the triangle inequality

$$\begin{aligned} \bigl|\varGamma_{2,\nu}^{k}(\mu) - \varGamma_2^{k}( \mu)\bigr| \leq C_T \bigl(1+I(\mu|P) \bigr) d_T(\mu,\nu). \end{aligned}$$

Proof of Lemma 5.(v)

For all α∈{1⋯M}, let

$$\begin{aligned} dQ_{\nu}^{\alpha, k}(x) = \exp{\varGamma_{\nu}^{\alpha ,k}( \delta_x)}dP_{\alpha} (x) = &\int\exp \biggl( \int_0^T \bigl(G^{\alpha}_{t^{(k)}} + m^{\alpha }_{\nu} \bigl(t^{(k)} \bigr)\bigr) \, dW^{\alpha}_t(x) \\ &{}- \frac{1}{2} \int_0^T \bigl(G^{\alpha}_{t^{(k)}} + m^{\alpha}_{\nu} \bigl(t^{(k)}\bigr)\bigr)^2 dt \biggr) \, d \gamma_{\nu} \, dP_{\alpha}(x). \end{aligned}$$

The equality between the two expression of $Q_{\nu}^{\alpha,k}$ is easily obtained by Gaussian calculus (see the proof of Proposition 4). We deduce by the martingale property of this density that it is a probability measure on $\mathcal{C}([-\tau ,T],\mathbb{R})$.

Remark that

It follows that $Q_{\nu}^{k} \in\mathcal{M}_{1}^{+}(\mathcal{C})$, and

$$\begin{aligned} dQ_{\nu}^k(x) = \exp{\varGamma_{\nu}^{k}( \delta_x)}dP(x) = &\int\exp \biggl( \int_0^T \bigl( \mathbf{G}_{t^{(k)}} + \mathbf{m}_{\nu}\bigl(t^{(k)} \bigr) \bigr)' \cdot\, d\mathbf{W}_t(x) \\ &{} - \frac{1}{2} \int_0^T \bigl\| \mathbf {G}_{t^{(k)}} + \mathbf{m}_{\nu}\bigl(t^{(k)} \bigr) \bigr\|^2 dt \biggr) \, d\gamma_{\nu} \, dP(x). \end{aligned}$$

Lets now prove that $H^{k}_{\nu} =I(.|Q^{k}_{\nu})$. We will first show that $I(Q^{k}_{\nu}|P)$ is finite. In fact,

$$\begin{aligned} \frac {\mbox {d} Q_{\nu}^{\alpha, k}}{\mbox {d} P_{\alpha}} (x) = &\biggl(\int\exp \biggl\{ -\frac{1}{2} \int_0^T {G^{\alpha}_{t^{(k)}}}^2 + {m^{\alpha}_{\nu}}^2 \bigl(t^{(k)}\bigr) dt \biggr\} d\gamma_{\nu} \biggr) \exp \biggl\{\int_0^T m^{\alpha}_{\nu} \bigl(t^{(k)}\bigr) dW^{\alpha }_t(x) \biggr\} \\ &{}\times\exp\biggl\{\frac{1}{2} \int\biggl(\int_0^T G^{\alpha }_{t^{(k)}} \bigl( dW^{\alpha}_t(x) - m^{\alpha}_{\nu}\bigl(t^{(k)}\bigr) dt \bigr) \biggr)^2 d\gamma_{\widetilde{K}_{\nu}^{T,k}} \biggr\}. \end{aligned}$$

(43)

Which becomes, after some Gaussian computations (see [3, Lemma 5.15]):

$$\begin{aligned} \frac {\mbox {d} Q_{\nu}^{\alpha, k}}{\mbox {d} P_{\alpha}} = \exp\biggl\{\int _0^T H^{\alpha }_{t^{(k)} } \bigl(Q_{\nu}^{\alpha, k}\bigr)dW^{\alpha}_t- \frac{1}{2}\int_0^T {H^{\alpha }_{t^{(k)} }}^2 \bigl(Q_{\nu}^{\alpha, k}\bigr)dt \biggr\} \end{aligned}$$

where

$$\begin{aligned} H^{\alpha}_{t^{(k)}}\bigl(Q^{\alpha, k}_{\nu}\bigr) = & \biggl(\int G^{\alpha }_{t^{(k)}} \int_0^t G^{\alpha}_{s^{(k)}} \bigl(dW^{\alpha}_s - m^{\alpha}_{\nu }\bigl(s^{(k)} \bigr)ds \bigr) \; d \gamma_{\widetilde{K}_{\nu}^{t,k}} \biggr) + m^{\alpha }_{\nu } \bigl(t^{(k)}\bigr) \\ = &\int_0^t \bigl(\widetilde{K}_{\nu}^{t,k} \bigl(t^{(k)},s^{(k)}\bigr) \bigr)_{\alpha\alpha } \bigl(dW^{\alpha}_s - m^{\alpha}_{\nu} \bigl(s^{(k)}\bigr)ds \bigr) + m^{\alpha }_{\nu } \bigl(t^{(k)}\bigr) \\ = &\sum_{l=0,\ldots,k ; t_{l+1}\leq t} \bigl(W^{\alpha }_{t_{l+1}}-W^{\alpha }_{t_l} - m^{\alpha}_{\nu}(t_l) (t_{l+1}-t_l) \bigr) \bigl(\widetilde{K}_{\nu}^{t,k}\bigl(t^{(k)}, t_k\bigr) \bigr)_{\alpha\alpha} + m^{\alpha }_{\nu } \bigl(t^{(k)}\bigr). \end{aligned}$$

Hence, according to Girsanov Theorem, there exists a $Q_{\nu}^{\alpha, k}$-Brownian motion B ^α such that

$$\begin{aligned} W^{\alpha}_t=B^{\alpha}_t+\int _0^t H^{\alpha}_{s^{(k)}} \bigl(Q_{\nu }^{\alpha, k}\bigr)ds. \end{aligned}$$

As W is an affine function of B ^α, it is, under $Q_{\nu }^{\alpha, k}$, a Gaussian variable with finite moments. In particular, $I(Q_{\nu}^{\alpha, k}|P_{\alpha})$ is finite. But

$$\begin{aligned} I\bigl(Q_{\nu}^k|P\bigr)= \int_{\mathcal{C}} \log\biggl( \frac {\mbox {d} Q^k_{\nu }}{\mbox {d} P} (x) \biggr) dQ^k_{\nu}(x) =& \sum_{\alpha=1}^M \int_{\mathcal{C}([-\tau,T],\mathbb{R})} \log\biggl( \frac {\mbox {d} Q^{\alpha, k}_{\nu}}{\mbox {d} P_{\alpha}} \bigl(x^{\alpha}\bigr) \biggr) dQ^{\alpha, k}_{\nu } \bigl(x^{\alpha }\bigr) \\ =& \sum_{\alpha=1}^M I\bigl(Q_{\nu}^{\alpha, k}|P_{\alpha}\bigr), \end{aligned}$$

so that $I(Q_{\nu}^{k}|P)$ is finite.

Now, let

$$\begin{aligned} \biggl(\tau^{\alpha}_{m}(x)=\inf\biggl\{t \geq0 ; \biggl| \int _0^t m^{\alpha}_{\nu} \bigl(s^{(k)}\bigr) dW^{\alpha}_s(x) \biggr| \geq m \biggr\} \biggr)_{m \in \mathbb{N}} \end{aligned}$$

be a sequence of stopping times for the Brownian filtration $\sigma( W^{\alpha}_{s}, 0\leq s \leq T)$. As W ^α is a P _α -MB, we have $\lim_{m\to\infty} \tau ^{\alpha }_{m} \to\infty$ almost surely under P _α . $( \tau_{m}= \min_{\alpha=1 \cdots M} \tau^{\alpha}_{m} )_{m\in\mathbb{N} }$ defines a sequence stopping times for $\sigma( W^{\alpha}_{s}, \alpha =1\cdots M, 0\leq s \leq T)$ which tends to infinity along with m P a.s. We define:

$$\begin{aligned} Q_{\nu,t}^k=\int\exp\biggl\{ \biggl( \int _0^t \bigl(\mathbf{G}_{s^{(k)}} + \mathbf{m}_{\nu}\bigl(s^{(k)} \bigr)\bigr)' \cdot \, d\mathbf{W}_s - \frac{1}{2} \int_0^t \bigl\|\mathbf{G}_{s^{(k)}} + \mathbf{m}_{\nu}\bigl(s^{(k)} \bigr) \bigr\|^2 ds \biggr) \biggr\} \, d\gamma_{\nu} \, P, \end{aligned}$$

$$\begin{aligned} \varGamma^{k}_{\nu,t}(\mu) =& \int_{\mathcal{C}} \log\biggl( \int\exp\biggl\{ \int_0^t \bigl( \mathbf{G}_{s^{(k)}}(\omega)+\mathbf{m}_{\nu }\bigl(s^{(k)} \bigr) \bigr)' \cdot d\mathbf{W}_s(x) \\ &{}- \frac{1}{2} \int_0^t \bigl\| \mathbf{G}_{s^{(k)}}(\omega)+\mathbf{m}_{\nu } \bigl(s^{(k)}\bigr) \bigr\| ^2ds \biggr\} d\gamma_{K^t_{\nu}}( \omega) \biggr) d\mu(x), \end{aligned}$$

as well as $Q_{\nu,t}^{\alpha,k}$ and $\varGamma^{\alpha, k}_{\nu,t}$ where

$$\begin{aligned} K_{\mu}^t(s,u)= \Biggl( \mathbf{1}_{\alpha=\beta} \displaystyle{\frac {1}{\lambda_{\alpha}^2} \sum_{\gamma=1}^M \sigma_{\alpha\gamma }^2 \int_{\mathcal{C}} S_{\alpha\gamma}\bigl(x^{\gamma}_{s-\tau_{\alpha\gamma} }\bigr)S_{\alpha\gamma} \bigl(x^{\gamma}_{u-\tau_{\alpha\gamma}}\bigr) d\mu(x)} \Biggr)_{\alpha , \beta\in\{1 \cdots M\}}. \end{aligned}$$

is define on [0,t]².

These functions are clearly continuous in time on [0,T]. The result $I(Q_{\nu,\tau_{m}\wedge T}^{k}|P) < \infty$ obviously remains.

By Jensen inequality, we have

$$\begin{aligned} \frac {\mbox {d} Q_{\nu,\tau^{\alpha}_{m}\wedge T}^{\alpha, k}}{\mbox {d} P_{\alpha }} (x) \geq&\exp\biggl \{-\frac{k_{\alpha}(\tau^{\alpha}_{m}(x)\wedge T)}{2\lambda _{\alpha}^2} \biggr\} \exp\biggl\{-\frac{\bar{J}_{\alpha}^2 (\tau ^{\alpha}_{m}(x)\wedge T)}{2 \lambda_{\alpha}^2} \biggr\} \\ &{}\times \exp \biggl\{\int_0^{\tau^{\alpha}_{m}(x)\wedge T} m^{\alpha}_{\nu} \bigl(t^{(k)}\bigr) dW^{\alpha }_t(x) \biggr\} \\ \geq&\exp\biggl\{-\frac{(k_{\alpha}+ \bar{J}_{\alpha}^2) T}{2\lambda_{\alpha}^2}-m \biggr\}, \end{aligned}$$

so that

$$\begin{aligned} \frac {\mbox {d} Q_{\nu,\tau_{m}\wedge T}^k}{\mbox {d} P} (x) \geq\exp\Biggl\{-\sum _{\alpha =1}^M \frac{(k_{\alpha}+ \bar{J}_{\alpha}^2) T}{2\lambda_{\alpha }^2}-Mm \Biggr\}. \end{aligned}$$

We then apply the same proof as in [3, Appendix B] to find:

$$\begin{aligned} \forall\mu\in\mathcal{M}_1^+(\mathcal{C}), \ H_{\nu,\tau _{m}\wedge T}^k = I\bigl(\mu|Q_{ \nu,\tau_{m}\wedge T}^k \bigr). \end{aligned}$$

Letting m to infinity, we conclude using the continuity of $H_{\nu ,t}^{k}$ and $I(.|Q_{\nu,t}^{k})$ on [0,T].

Proof of Lemma 5.(vi)

In order to demonstrate that H ^k is a good rate function, we need to show that it is lower semi-continuous and that it has compact level sets, i.e. {H ^k≤L} is a compact set for any L>0. This is a direct consequence of points (i)–(iv) proved above. □

Appendix B: Large Deviation Principle: Proof of the Technical Lemma 2

This appendix is concerned with the proof of Lemma 2 ensuring an exponential bound that will be used to show a tightness result on the sequence of empirical laws.

Proof

Let

$$\begin{aligned} B^n = \int_{\hat{\mu}_n \in B(\nu,\delta)} \exp\bigl\{ a n \bigl( \varGamma(\hat{\mu}_n)- \varGamma_{\nu}(\hat{ \mu}_n) \bigr) \bigr\} dQ_{\nu}^n. \end{aligned}$$

We introduce the probability measure Q _ν defined on $\mathcal {C}$ by:

$$\begin{aligned} \frac{dQ_{\nu}}{dP}(x) = \exp\bigl(\varGamma_{\nu}( \delta_x)\bigr) =& \int\exp \biggl(\int_0^T \bigl(\mathbf{G}_t + \mathbf{m}_{\nu}(t) \bigr)' \cdot d\mathbf{W}_t(x) \\ &{} - \frac {1}{2}\int _0^T \bigl\|\mathbf{G}_t + \mathbf{m}_{\nu}(t) \bigr\|^2 dt \biggr) d\gamma_{\nu }, \end{aligned}$$

so that $Q_{\nu}^{n}=Q_{\nu}^{\otimes n}$. Writing the definitions of Γ and Γ _ν , we find:

$$\begin{aligned} B^n = \int_{\hat{\mu}_n \in B(\nu,\delta)} \prod _{i=1}^n \biggl( \frac {{\mathcal{E}}_{\hat{\mu}_n} [\exp\{ \int_0^T ( \mathbf{G}_t + \mathbf{m}_{\hat {\mu}_n}(t) )' \cdot d\mathbf{W}^i_t -\frac{1}{2} \int_0^T \| \mathbf{G} _t + \mathbf{m} _{\hat{\mu}_n}(t) \|^2 dt \} ]}{{\mathcal{E}}_{\nu } [\exp \{ \int_0^T ( \mathbf{G}_t + \mathbf{m}_{\nu}(t) )' \cdot d\mathbf{W} ^i_t - \frac {1}{2} \int_0^T \| \mathbf{G}_t + \mathbf{m}_{\nu}(t) \| ^2 dt \} ]} \biggr)^a d(Q_{\nu})^{\otimes n}. \end{aligned}$$

Let ξ be a probability measure on $\mathcal{C}\times\mathcal {C}$ with marginals $\hat{\mu}_{n}$ and ν. We then have:

$$\begin{aligned} B^n = \int_{\hat{\mu}_n \in B(\nu,\delta)} \prod _{i=1}^n \biggl( \frac {{\mathcal{E}}_{\xi} [\exp\{ \int_0^T ( \mathbf {G}_t + \mathbf {m}_{\hat{\mu }_n}(t) )' \cdot d\mathbf{W}^i_t - \frac{1}{2} \int_0^T \| \mathbf{G}_t + \mathbf{m} _{\hat{\mu}_n}(t) \|^2 dt \} ]}{{\mathcal{E}}_{\xi } [\exp \{ \int_0^T ( \mathbf{G}_t' + \mathbf{m}_{\nu}(t) )d\mathbf{W}^i_t - \frac{1}{2} \int_0^T \| \mathbf{G}_t' + \mathbf{m}_{\nu}(t) \|^2 dt \} ]} \biggr)^a d(Q_{\nu})^{\otimes n} \end{aligned}$$

where (G,G′) is a 2M-dimensional Gaussian centered process with covariance K _ξ (see (34)).

$$\begin{aligned} K_{\xi}(s,t) = \left ( \begin{array}{c@{\quad}c} K_{\mu}(s,t) & (\mathbf{1}_{\{\alpha=\gamma\}} K^{\alpha }_{\xi }(s,t) )_{\alpha, \gamma= 1 \cdots M} \\ (\mathbf{1}_{\{\alpha=\gamma\}} K^{\alpha}_{\xi}(s,t) )_{\alpha, \gamma= 1 \cdots M} & K_{\nu}(s,t) \\ \end{array} \right ) . \end{aligned}$$

(44)

Let

Then

$$\begin{aligned} B^n = &\int_{\hat{\mu}_n \in B(\nu,\delta)} \prod _{i=1}^n \biggl( \frac {{\mathcal{E}}_{\xi} [\exp\{ Y_i \} ]}{{\mathcal {E}}_{\xi} [\exp \{ Y_i' \} ]} \biggr)^a d(Q_{\nu})^{\otimes n} \\ = &\int_{\hat{\mu}_n \in B(\nu,\delta)} \prod_{i=1}^n \biggl( {\mathcal{E}}_{\xi } \biggl[ \frac{\exp{Y_i'}}{{\mathcal{E}}_{\xi} [\exp{Y_i'} ]} \exp{ \bigl( Y_i - Y_i' \bigr) } \biggr] \biggr)^a d(Q_{\nu})^{\otimes n} \\ \leq&\int_{\hat{\mu}_n \in B(\nu,\delta)} \prod_{i=1}^n {\mathcal{E} }_{\xi} \biggl[ \frac{\exp{Y_i'}}{{\mathcal{E}}_{\xi} [\exp { Y_i'} ]} \exp{ a \bigl( Y_i - Y_i' \bigr) } \biggr] d(Q_{\nu})^{\otimes n} \end{aligned}$$

by Jensen inequality.

Then, using Holder inequality twice with conjugate exponents (p,q) and (σ,η), one finds:

$$\begin{aligned} B^n \leq&\Biggl\{ \overbrace{\int\prod_{i=1}^n \frac{{\mathcal {E}}_{\xi} [\exp{ pY_i'} ]}{ ( {\mathcal{E}}_{\xi} [\exp{Y_i'} ] )^p } d(Q_{\nu})^{\otimes n}}^{B^n_1} \Biggr \}^{\frac{1}{p}} \biggl\{ \overbrace{\int\exp\bigl\{n \sigma \varGamma_{\nu}(\hat{\mu}_n) \bigr\} dP^{\otimes n}}^{B^n_2} \biggr\}^{\frac{1}{q\sigma}} \\ &{}\times\Biggl\{ \underbrace{\int_{\hat{\mu}_n \in B(\nu,\delta)} \prod _{i=1}^n {\mathcal{E}}_{\xi} \bigl[ \exp{ a \eta q \bigl( Y_i -Y_i' \bigr) } \bigr] dP^{\otimes n}}_{B^n_3} \Biggr\}^{\frac{1}{q\eta}}. \end{aligned}$$

(45)

We first bound the first term of the right hand side. Let $\gamma _{p,\widetilde{K}^{T}_{\mu}}$ be a probability measure on Ω such that $d\gamma_{p,\widetilde{K}^{T}_{\mu}} = \frac{\prod_{\gamma =1}^{M} \exp {-\frac{p}{2}\int_{0}^{T} {G^{\gamma}_{t}}^{2} dt} }{\int\prod_{\gamma=1}^{M} \exp{-\frac{p}{2}\int_{0}^{T} {G^{\gamma}_{t}}^{2} dt} d\gamma_{\mu} } d\gamma _{\mu}$. According to appendix A of [3] (where p=β ²), we have, for any p≥0, that G is a M-dimensional centered Gaussian process under $\gamma_{p,\widetilde {K}^{T}_{\mu}}$.

Consequently, using the independence of (G,G′)’s components:

$$\begin{aligned} &{\mathcal{E}}_{\xi} \bigl[\exp{ pY_i'} \bigr] \\ &\quad{} = \prod_{\alpha =1}^M \exp\biggl\{ p \biggl( \int_0^T m_{\nu}^{\alpha}(t) dW^{i_{\alpha}}_t - \frac {1}{2} \int_0^T {m^{\alpha}_{\nu}}^2(t) dt \biggr) \biggr\} \; { \mathcal{E}}_{\xi} \biggl[ \exp\biggl\{-\frac{p}{2} \int _0^T {{G^{\alpha}_t}'}^2 dt \biggr\} \biggr] \\ &\qquad{}\times\int\exp\biggl\{ p \biggl(\int_0^T {G^{\alpha}_t}' \bigl(dW^{i_{\alpha}}_t - m_{\nu}^{\alpha}(t) dt \bigr) \biggr) \biggr\} d\gamma _{p,\widetilde{K}^T_{\nu}} \\ &\quad{} = \prod_{\alpha=1}^M \exp\biggl\{ p \biggl( \int_0^T m_{\nu }^{\alpha}(t) dW^{i_{\alpha}}_t - \frac{1}{2} \int_0^T {m^{\alpha}_{\nu}}^2(t) dt \biggr) \biggr\} \; { \mathcal{E}}_{\xi} \biggl[ \exp\biggl\{-\frac{p}{2} \int _0^T {{G^{\alpha}_t}'}^2 dt \biggr\} \biggr] \\ &\qquad{}\times \exp\biggl\{ \biggl( \frac{p^2}{2} \int\biggl(\int _0^T G^{\alpha }_t \bigl(dW^{i_{\alpha}}_t - m_{\nu}^{\alpha}(t) dt \bigr) \biggr)^2 \frac {\exp { (-\frac{p}{2}\int_0^T {G^{\alpha}_t}^2 dt )} }{\int \exp { (-\frac{p}{2}\int_0^T {G^{\alpha}_t}^2 dt )} d\gamma _{\nu}} d\gamma_{\nu} \biggr) \biggr\} . \end{aligned}$$

Hence,

$$\begin{aligned} \frac{{\mathcal{E}}_{\xi} [ \exp{pY_i'} ]}{{\mathcal {E}}_{\xi} [ \exp {Y_i'} ]^p} = &\prod_{\alpha=1}^M \underbrace{\frac{{\mathcal{E}}_{\xi } [ \exp {-\frac{p}{2} \int_0^T {{G^{\alpha}_t}'}^2 dt} ]}{{\mathcal {E}}_{\xi } [ \exp{-\frac{1}{2} \int_0^T {{G^{\alpha}_t}'}^2 dt} ]^p}}_{f^{\alpha }(p)} \exp\biggl\{ \frac{p}{2} \int\biggl(\int_0^T G^{\alpha}_t \bigl(dW^{i_{\alpha}}_t - m_{\nu}^{\alpha}(t) dt \bigr) \biggr)^2 \\ &{} \times\underbrace{ \biggl(p\frac{\exp{-\frac{p}{2}\int_0^T {G^{\alpha }_t}^2dt} }{\int\exp{-\frac{p}{2}\int_0^T {G^{\alpha}_t}^2dt} d\gamma _{\nu}} - \varLambda_T \bigl(G^{\alpha}\bigr) \biggr)}_{g^{\alpha}(p)} d\gamma_{\nu} \biggr\}. \end{aligned}$$

Remark thatf ^α(p) is bounded for p>0. In fact, on one hand it is clear that

$$\begin{aligned} \forall p \in[0,+\infty[, {\mathcal{E}}_{\xi} \biggl[ \exp{- \frac {p}{2} \int_0^T {{G^{\alpha}_t}'}^2 dt} \biggr]\leq1, \end{aligned}$$

furthermore Jensen inequality and Fubini theorem give us:

$$\begin{aligned} {\mathcal{E}}_{\nu} \biggl[\exp\biggl\{-\frac{p}{2}\int _0^T {G^{\alpha}_t}^2 dt \biggr\} \biggr] \geq\exp\biggl\{ - \frac{p}{2} \int _0^T {\mathcal{E}}_{\nu } \bigl[ {G^{\alpha }_t}^2 \bigr] dt \biggr\} \geq\exp\biggl \{ -\frac{p T k_{\alpha}}{2 \lambda_{\alpha}^2} \biggr\} . \end{aligned}$$

Therefore, bounded convergence monotone gives f ^α(p)→1 and, similarly, g ^α(p)→0 as p↘1. Moreover, as $(\int_{0}^{T} {G^{\alpha}_{t}}' (dW^{i_{\alpha}}_{t} - m_{\nu}^{\alpha}(t) dt ) )^{2}$ has finite moments under γ _ν , we can find a finite constant C ₁(p), C ₁(p)↘0 as p↘1, such that:

$$\begin{aligned} B^n_1 = \biggl( \int\frac{{\mathcal{E}}_{\xi} [\exp{ pY_i'} ]}{ ( {\mathcal{E} }_{\xi} [\exp{ Y_i'} ] )^p } dQ_{\nu} \biggr)^n \leq e^{ C_1(p) n} . \end{aligned}$$

(46)

Moreover:

$$\begin{aligned} B^n_2 = & \biggl( \int{\mathcal{E}}_{\nu} \biggl[ \exp\biggl\{ \int_0^T \bigl( \mathbf{G}_t + \mathbf{m}_{\nu}(t) \bigr)' \cdot d\mathbf{W}_t(x) - \frac{1}{2} \int_0^T \bigl\|\mathbf{G}_t + \mathbf{m}_{\nu}(t) \bigr\|^2 dt \biggr\} \biggr]^{\sigma} dP(x) \biggr)^n \\ \leq&\biggl( {\mathcal{E}}_{\nu} \biggl[ \int\exp\biggl\{ \sigma \int_0^T \bigl(\mathbf{G} _t + \mathbf{m}_{\nu}(t) \bigr)' \cdot d\mathbf{W}_t(x) \biggr\} dP(x)\\ &{}\times \exp\biggl\{ - \frac{\sigma}{2} \int_0^T \bigl\|\mathbf{G}_t + \mathbf{m}_{\nu }(t) \bigr\|^2 dt \biggr\} \biggr] \biggr)^n \\ = & \biggl( {\mathcal{E}}_{\nu} \biggl[ \exp\biggl\{ \frac{\sigma ^2-\sigma }{2} \int_0^T \bigl\| \mathbf{G}_t + \mathbf{m}_{\nu}(t) \bigr\|^2 dt \biggr\} \biggr] \biggr)^n. \end{aligned}$$

So that if we take σ close enough to 1, we can find a finite constant C ₂(σ), lim _σ→1 C ₂(σ)=0, such that (see inequality (38)):

$$\begin{aligned} B^n_2 \leq\biggl( {\mathcal{E}}_{\nu} \biggl[ \exp\biggl\{ \frac {\sigma ^2-\sigma }{2} \int_0^T \bigl\| \mathbf{G}_t + \mathbf{m}_{\nu}(t) \bigr\|^2 dt \biggr\} \biggr] \biggr)^n < e^{ C_2(\sigma) n } . \end{aligned}$$

(47)

We now will bound the last term of the right hand side of (45). By Cauchy-Schwarz inequality, if κ=qaη:

$$\begin{aligned} B^n_3 \leq&\Biggl\{ \int_{\hat{\mu}_n \in B(\nu,\delta)} \prod_{i=1}^n {\mathcal{E}}_{\xi} \biggl[ \exp\biggl\{ 2\kappa\int_0^T \bigl( \mathbf{G}_t-\mathbf{G} _t'+ (\mathbf{m} _{\hat{\mu}_n}-\mathbf{m}_{\nu}) (t) \bigr)' \cdot d \mathbf{W}^i_t \\ &{}- 2\kappa^2 \int_0^T \bigl \| \mathbf{G}_t - \mathbf{G}_t' + (\mathbf{m}_{\hat {\mu}_n} -\mathbf{m}_{\nu}) (t) \bigr\|^2 dt \biggr\} \biggr] dP^{\otimes n} \Biggr\} ^{\frac{1}{2}} \\ &{}\times \biggl\{ \int_{\hat{\mu}_n \in B(\nu,\delta )} {\mathcal{E} }_{\xi} \biggl[ \exp \biggl\{ 2\kappa^2 \int_0^T \bigl\| \mathbf{G}_t - \mathbf{G}_t' + ( \mathbf{m}_{\hat{\mu}_n} - \mathbf{m}_{\nu }) (t) \bigr\|^2 dt \\ &{} - \kappa\int_0^T \bigl\|\mathbf{G}_t+ \mathbf{m}_{\hat{\mu }_n}(t) \bigr\| ^2 - \bigl\| \mathbf{G} _t' + \mathbf{m}_{\nu}(t) \bigr\|^2 dt \biggr\} \biggr]^n dP^{\otimes n} \biggr\} ^{\frac{1}{2}}. \end{aligned}$$

The first term is the square root of a martingale’s expectation, thus equal to one. For the second term, we remark that:

$$\begin{aligned} &- \int_0^T \bigl(G^{\alpha}_t+m^{\alpha}_{\hat{\mu}_N}(t) \bigr)^2 - \bigl({G^{\alpha}_t}' + m_{\nu}^{\alpha}(t) \bigr)^2 dt\\ &\quad{} \leq \frac{\delta ^{\frac{1}{2}}}{2} \biggl( \frac{1}{\delta} \int_0^T \bigl(G^{\alpha}_t- {G^{\alpha}_t}' + \bigl(m^{\alpha}_{\hat{\mu}_N}(t) - m_{\nu}^{\alpha } \bigr) (t) \bigr)^2 dt \\ &\qquad{}+ \int_0^T \bigl(G^{\alpha}_t+ {G^{\alpha}_t}' + \bigl(m^{\alpha}_{\hat {\mu }_N}(t)+m^{\alpha}_{\nu} \bigr) (t) \bigr)^2 dt \biggr), \end{aligned}$$

so that, by Cauchy-Schwarz inequality,

$$\begin{aligned} B^n_3 \leq&\biggl\{ \int{\mathcal{E}}_{\xi} \biggl[ \exp\biggl\{ \bigl(4\kappa^2+\kappa\delta^{-\frac{1}{2}} \bigr) \int_0^T \bigl\| \mathbf{G}_t- \mathbf{G} _t'+ (\mathbf{m} _{\hat{\mu}_n}- \mathbf{m}_{\nu}) \bigl(t^{(k)}\bigr) \bigr\|^2 dt \biggr\} \biggr]^n dP^{\otimes n} \biggr\}^{\frac{1}{4}} \\ &{} \times\biggl\{ \int{\mathcal{E}}_{\xi} \biggl[ \exp\biggl\{ \kappa \delta^{\frac {1}{2}}\int_0^T \bigl\| \mathbf{G}_t+\mathbf{G}_t'+ (\mathbf {m}_{\hat{\mu }_n}+\mathbf{m}_{\nu}) (t) \bigr\|^2 dt \biggr\} \biggr]^n dP^{\otimes n} \biggr\}^{\frac{1}{4}} \\ \leq&\exp{\sum_{\alpha=1}^n \biggl\{ \frac{1}{2} \bigl(4\kappa^2+\kappa\delta^{-\frac{1}{2}} \bigr)\frac{\bar{J}_{\alpha}^2 K_S^2}{\lambda_{\alpha }^2}T\delta^2 + \biggl(2\kappa\delta ^{\frac{1}{2}}T\frac{\bar{J}_{\alpha}^2}{\lambda_{\alpha}^2}\biggr) \biggr\}} \\ &{} \times\biggl\{ \int{\mathcal{E}}_{\xi} \biggl[ \exp\biggl\{ 2 \bigl(4\kappa^2+\kappa\delta^{-\frac{1}{2}} \bigr) \int _0^T \|\mathbf{G}_t-\mathbf{G} _t' \|^2 dt \biggr\} \biggr]^n dP^{\otimes n} \biggr\}^{\frac{1}{4}} \\ & {}\times\biggl\{ \int{\mathcal{E}}_{\xi} \biggl[ \exp\biggl\{ 2 \kappa\delta^{\frac {1}{2}}\int_0^T \| \mathbf{G}_t+\mathbf{G}_t' \|^2 dt \biggr\} \biggr]^n dP^{\otimes n} \biggr\}^{\frac{1}{4}}. \end{aligned}$$

As ${\mathcal{E}}_{\xi} [ \int_{0}^{T} \|\mathbf{G}_{t}-\mathbf {G}_{t}' \|^{2} dt ] = \sum_{\alpha, \gamma=1}^{M} \frac{\sigma_{\alpha\gamma}^{2} K_{S}^{2}}{\lambda_{\alpha} ^{2}} \int_{0}^{T} \int| S_{\alpha\gamma}(x^{\gamma}_{t-\tau_{\alpha\gamma}}) - S_{\alpha \gamma}(y^{\gamma}_{t-\tau_{\alpha\gamma}}) |^{2} d\xi (x,y) dt \leq\sum_{\alpha, \gamma=1}^{M} \frac{\sigma_{\alpha \gamma}^{2} T}{\lambda_{\alpha}^{2}} \delta ^{2}$ (in fact, $d_{T}(\hat{\mu}_{n},\nu) \leq\delta$), we can use Appendix A, Lemma 3.2 of [3] to bound the two last term of the previous inequality: there exists two finite constants $C^{\kappa }_{1}(\delta)$ and $C^{\kappa}_{2}(\delta)$ such that

Hence, we can find C _κ (δ),lim _δ→0 C _κ (δ)=0, such that

$$\begin{aligned} B^n_3 \leq e^{C_{\kappa}(\delta) n} . \end{aligned}$$

(48)

We conclude by using (46), (47) and (48) in (45). □